[Math] Should BODMAS not be BODMSA

arithmetic

Let me blindly follow for a second BODMAS.
$$
\begin{align*}
1-3+2
&=1-(3+2)\\
&=1-5\\
&=-4
\end{align*}
$$
(The brackets are just to make the error clear – I wouldn't write them in "real life".)
This is correct according to BODMAS, because BODMAS doesn't talk about positive or negative numbers. It just says that addition should come before subtraction.

What am I missing here? Is it like "i before e except after c", and then you realise their is a word and the whole rhyme is as follows:

i before e,

Except after c,

Or when sounded as "a,"

As in neighbour and weigh.

So, am I missing some subtlety? Is the rhyme incomplete? Is it not true that subtraction has greater precedence than addition, and so BODMSA is actually correct and BODMAS is incorrect?

NOTE: This issue is true for all other mnemonics I know of, such as BIDMAS, PEMDAS, PIDMAS…

Best Answer

In the Danish educational system we are taught the order of operations in the following hierarchy:

  1. Brackets
  2. Roots and powers
  3. Multiplication and division
  4. Addition and subtraction

along with the rule of carrying out operations of equal order in the reading direction. In effect we have a BO(D,M)(A,S) system where (D,M) and (A,S) are pairs of equal order. This yields the same results for any calculation as all international standards prescribe, so no need to worry. The funny thing is that in our system you are meant to calculate as follows

$$ a\cdot b/c=(a\cdot b)/c $$ whereas the English BODM(A,S) prescribes $$ a\cdot b/c=a\cdot(b/c) $$ Of course those two expressions are equal to one another, so it produces no controversies about Danish and international order of operation standards.

My best guess is that BODMAS was ordered that way in order to form a pronouncable word rather than the tongue twisting BODMSA knowing that A and S are equal as long as you apply the rule of the reading direction too.

NOTE: Actually PEMDAS would cause problems too unless we interpreted it as PE(M,D)(A,S). For instance $a/b\cdot c$ would equal $a/(b\cdot c)$ if multiplication really came before division. But in reality what is meant is reading from the left to the right performing operations of equal order, namely $a/b\cdot c=(a/b)\cdot c$.

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